We have the equation \begin{equation} f(\mathbf{x})=\mathbf{c}^T\mathbf{x}-\sum_{i=1}^m\log{(b_i-\mathbf{a}_i^T\mathbf{x})} ,\;\;\; \mathbf{x} \in \mathbb{R}^n \text{ and } m>n \end{equation}
where we see that $\textbf{dom}f$ is convex since it is the a halfspace of the hyperplane $\mathbf{a}^T\mathbf{x}=b$.
I want to show that $f$ is convex. What is the best approach to do so? Should I differentiate or should I show that Jensen's inequality holds truth?
I tried with Jensen's inequality and for $\mathbf{x}_1,\mathbf{x}_2 \in$ $\textbf{dom}f$ and $\theta \in [0,1]$ I begin by \begin{equation} f(\theta \mathbf{x}_1+(1-\theta)\mathbf{x}_2)<\theta f(\mathbf{x}_1)+(1-\theta)\mathbf{x}_2 \end{equation} and at some point I get: \begin{equation} \begin{aligned} -\sum_{i=1}^m\log{(b_i-\mathbf{a}_i^T(\theta \mathbf{x}_1+(1-\theta)\mathbf{x}_2))}<&-\theta \sum_{i=1}^m\log{(b_i-\mathbf{a}_i^T\mathbf{x}_1)}\\&- \sum_{i=1}^m\log{(b_i-\mathbf{a}_i^T\mathbf{x}_2)}\\&+\theta \sum_{i=1}^m\log{(b_i-\mathbf{a}_i^T\mathbf{x}_2)} \end{aligned} \end{equation}
From this point forward how can I proceed with Jensen's inequality to show that f is convex?